Higher Order Derivatives Week2-Degree

8/14/2019 Higher Order Derivatives Week2-Degree

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HIGHER ORDER DERIVATIVES

.

1The second order partial derivatives

are found by differentiating the first

order partial derivatives

ii) Notation : 2

2

x

f

y x

f

2

x y

f

22

2

y

f


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y

=

2

2

) x

f

i

x

f Differentiatetwice with respectto x x

y

f Differentiatetwice with respectto y

=

2

2

) y

f ii

y

STEP 2 STEP 1

STEP 2 STEP 1


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=

y x

f i

2

)

y

f Differentiate firstwith respect to yand then withrespect to x x

x

f Differentiate firstwith respect to xand then withrespect to y

=

x y

f ii

2

) y

STEP 2 STEP 1

STEP 2 STEP 1


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EXAMPLE 1 : 2223 33),( y xy y x x y x f ++=Determine :

2

2

) x

f i

=

x f ( )2223 33 y xy y x x ++

Step 1 : find ,let y constant x f

x

y x

f

ii 2

)

Solution: i) To find =

2

2

x

f

x

f

xSTEP 1


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( ) ( ) ( ) (2223 33 y x

xy x

y x x

x x x

f +

+

=

( ) ( ) ( ) ( )2223

33 y x x x y x x y x x

+

+

=

Take out the coefficients

( ) ( )01323322

++= y x y x22 363 y xy x

x

f +=

constant

First order partialderivative is


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Therefore, the second partial derivatives is

= x f

x x f 2

2

( )22 363 y xy x +

( ) ( ) ( )22

363 y x x x y x x

+

=( ) ( ) ( )01623 += y x y x 66 =

Answer from step 1

constant

x

=


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Solution: i) Short-cut

2223

33),(Given y xy y x x y x f ++=1 st order : )0()1(3)2(33 22 ++=

y y x x

x

f

22 363 y xy x +=

2nd order : 0)1(6622

+=

y x x f

y x 66 =


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Solution: ii) To find x y f

2

=

x f

y STEP 1

Therefore, the second order partial derivatives is

=

x

f

y x y

f 2

( )22

363 yxyx +

Answer from step 1 (part i)

y

=

From part (i),first partial derivatives is 22 363 y xy x

x f +=

=


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( ) ( ) ( )22 363 y y

y y

x x y

+

=

( ) ( ) y x 2316)0( +=

yx 66 +=

constant

y x y f

=2

( )22

363 y xy x +Apply sum and different formula and take out the coefficients


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x y

f

y x

f

=

22

Equality of Mixed Second Order Partial Derivative Theorem


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EXAMPLE2:Given

2

),( y xe y x f =Verify (prove/show) that

x y f

y x f

=22

Solution : From the left hand side

= y x f 2

STEP 1 y f

x


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= y f ( )2 xye

Step 1 : find first partial derivativelet x constant

y f

y

( )2 xye=

( ) [ ]

= 22 y y

xe xy

[ ]

2 xy y

Take out the coefficients x

( ) [ ]( ) y xe xy 22=

Recalling that :

[ ] )('.)()( x f eedxd x f x f =

2

2 xy xye=


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= y f x y x f 2

( )2

2 xy xye x=

Answer from step 1

Take out the coefficients 2y

( )22 xy xex

y

= Product Rule



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( )222

xy xe x

y y x

f

=

[ ]''2 uvvu y +=

Factorizecommon

factor

[ ]))()(())(1(2 222 ye xe y xy xy +=

( )[ ]2122 xye y xy +=

( )2

122

xyye xy

+=

2 xye

Product rule


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SOLUTION : CONTINUE

Find the right hand side

=

x

f

y x y

f 2

STEP 1

x y f

2


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= x f ( )2 xye

Step 1 : find first partial derivativelet x constant

x f

x

( )2 xye=

( ) [ ]

= x x

ye xy 22

[ ]

2 xy x


( )( ))1(22 ye xy= 22 xye y=

[ ] )('.)()( x f eedx

d x f x f =


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=

x

f

y x y

f 2

( )22 xye y y=Answer from step 1

Product Rule



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( )222

xye y

y x y

f

=

[ ]'' uvvu +=[ ] [ ] + =

22222

)()( xy y

e ye y y

xy xy

[ ]

+=22 22

)())(2( y y xe ye yxy xy

)2())(())(2(22 2 y xe ye y xy xy +=

Product rule


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Factorizecommonfactor ( )212 2 xy ye xy +=

22 322 xy xy e xy ye +=2

2 xy ye

Hence :

x y f

y x f

= 22

( )2122

xy ye xy +=Verify !


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EXAMPLE 3 : Given

= x y

y x f ln),(

, evaluate22

2

2

2

2

y x f

y f

x f at 1,2 == y x

Solution:

= x y

y x f ln),( rewrite x y y x f lnln),( =

The first partial derivatives are :

xx

f 1=

y y

f 1=


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So, the second partial derivatives are :

=

x f

x x f 2

2

=

y f

x y x f 2

= x x1 ( )1

= x x 2

1 x

=

=

y

f

y y

f 2

2

= y y

1 ( )1

= y y

2

1

y

=

Answer of first order partial derivative


= y x1

0=



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Therefore given : 22

2

2

2

2

y x f

y f

x f

( )222 0

11

=

y x

= 22 1

121

1,2 == y x

( )141

=

when

41=


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Solution: Continue.

For this example, in stead of find y x z

2

We will find x y z

2because we can get the same answer quickly

=

x z

y x y z 2

STEP 1


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( )

++

=

12 ye

x xy

x x z y

( )

++

=

12 y

e

x x

x y

y

( ) 01 += y

constant

y x z =

Therefore,

= x

z y x y

z 2

( ) y y=

1=


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Remark : If we differentiate first we respect toy, we have to use quotient rule and

we are for significantly more work !

= y z

++

12 ye

xy y

y

Apply sum and different formula

( )

+

+

= 12 ye

y xy y

y

Quotient rule

Homework !


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More Higher Order Derivatives(3 rd Order, 4 th Order. )

EXAMPLE 5 : Let ye y y x f x += 2),(find

x y f

2

3

Solution: The formula

=

x y

f 2

3 Step 1 :diff. w.r. to x

Step 2 :diff. w.r. to y

Step 3 :diff. w.r. to y again


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=

x

f

ye y y x f x += 2),(Solution : GivenCopy coefficient

02 + xe y

constant

Step 2 : 2 nd order

(diff. w.r.to y)=

x y f 2 xe y2

xe y2=Differentiate w.r.to y

Step 1 : 1 st order

(diff. w.r.to x)

Step 3 : 3 rd order

(diff. w.r.to y again) =

x y f

2

3 xe)1(2 xe2=

Copy coefficient


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Chain Rule for functions of multiple intermediatevariables and 1 independent variable

If ),( y x f w = is differentiable and x and y are

t

dt dy

yw

dt dx

xw

dt dw

+

=

Remark : i) Notice that both partial derivative xw

and single-variable derivatives

dt

dx

are involved.

differentiable functions of t, then w is a differentiablefunction of t and :


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Example 1 :Suppose that y xw 2= where2t x = 3t y =

Use the chain rule to find dt dw

and check the result by expressing w

as a function of t and differentiatingdirectly.

and


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Solution : By using the chainrule

dt dy

yw

dt dx

xw

dt dw

+

=

)()()()(3222

t dt d

y x yt dt d

y x x

+

=

)3)(1.()2)(2( 22 t xt xy +=

)3)(()2)(.2( 2432 t t t t t +=

66

34 t t +=6

7t =

substitut e

y xw 2=2t x =3t y =

Given

Formula :

Answer in termof t

Solution :


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Solution :Continue.

method 2: We can express wdirectly as a function of t

y xw 2=S

o,

Remark : However, this procedure maynot

always be convenient

)()( 322 t t =7t =))(( 34 t t =

67t

dt

dw =

7

t w =substit

ute


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Example 2 : Suppose that y xy z +=

and,cos = x

,

sin= yUse the chain rule to find

d dz when

2

=


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Solution : By using the chain rule

d

dy

y

z

d

dx

x

z

d

dz

+

=

)(sin)()(cos)( dt

d y xy

ydt

d y xy

x+

++

=

substitut e

y xy z += cos= x sin= yGiven

Formula :

( ) ( ) ))(cos11.(21

)sin)(0.1(21

21

21

+++++= x y xy y y xy

Chainrule


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( ) ( ) ))(cos1(21

)sin)((21

21

21

++++=

x y xy y y xyd dz

y xy

x

y xy

y

+

++

+

=2

))(cos1(

2

)sin)((

y xy

x y

+++=

2

))(cos1()sin)((

sin))(sin(cos2

))(cos1(cos)sin)((sin

+++=

Solution : Continue

simplify

Combine the 2 terms

Answer interms of


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When 2

=

, 2sin)

2)(sin

2(cos2

)2)(cos12(cos)2sin)(2(sin

+

++=d dz

1)1)(0(2

)0)(10()1)(1(

+++=

d dz

21

102

01 =

+

+= d

dz


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Method 2 : If we express z directly as a

, we

obtain :

y xy z += cos= x sin= y

sin))(sin(cos += z

To get d

dz , we have to use product rule

function of

and it involve long working step !

substitute


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Case 2 : Given 3 intermediatevariables

),,( z y x f w = is differentiable and

dt dz

z w

dt dy

yw

dt dx

xw

dt dw

++=

x, y, z and 1 independent

variable t .

If x ,y and z are differentiable functions of t,

CHAIN

RULE :


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Example : Given z xyw += where,cos t x = ,sin t y = t z =

Finddt dw when 0=t

Solution:

dt dz

z w

dt dy

yw

dt dx

xw

dt dw

+

+=

)()()(sin)()(cos)( t dt d

z xy z

t dt d

z xy y

t dt d

z xy x

+++

++=

)1)(1())(cos01.()sin)(0.1( ++++= t xt y

substitut

e


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1))(cos()sin)(( ++= t xt y1))(cos(cos)sin)((sin ++= t t t t

1)(cos)(sin22

++= t t So, when 0=t

1)0(cos)0(sin22

++= 1)1()0( 22 ++=2

=

Solution : Continue

simplify

Answer in term of t


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A FORMULA FOR IMPLICITDIFFERENTIATION

),( y x F

and the equation 0),( = y x F

function of x. Then, at any point where0

y F , we obtain that :

y F

x F

dxdy

=

Case 1 :

Supposeis differentiable

defines y implicitly as a differentiable


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Example 1:

722 =+ y xy x, find dx

dy by using formula for implicit

result using implicitdifferentiation.

differentiation and checkthe

Given


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07),(Let22

=+= y xy x y x F Step 2 : By using formula for implicit differentiation,

y F

x F

dxdy

=

Solution : Given

Step 1: 0),( = y x F

722 =+ y xy x

Must bezero


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y F

x F

dxdy

=Formula

7),(Let22

+= y xy x y x F

Solution : Continue

00)1(2 ++= y x y x =2

02)1(0 ++=

y x

y

F y x 2+=

y x y x2

2+=

x y x y

=

22

simplify

x F


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Method 2: differentiating the given equation

722

=+ y xy x ,implicitly)7()( 22

dxd

y xy xdxd =+

x y

x y

dx

dy

=2

2

Remark: The calculation which using formula for implicit differentiation is significantly shorter than thesingle-variable calculation which we learned in Diploma

course.

Solve it by your-self


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Now supposethat z is given implicitly as

),( y x f z = by an equation of the form0),,( = z y x F . Then, at any point where

0 z F , we obtain

that :

z F

x F

x z

=

and z

F y F

y z

=

Case 2: For 3 variables x, y, z

a function of


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Example 2 : Given 223 333 =+++ xyz z y xfind

dx

z and

dy

z

Solution:The given equation can be written as

0223),,( 333 =+++= xyz z y x z y x F By using formula for implicit differentiation,

z F

x F

x z

=

y F

z =

Must bezero

03000009

2

2

++++++

= xy z yz x

xy z yz x

++

= 22

39

xz y += 226

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Higher Order Derivatives Week2-Degree